We have critical points anywhere the first derivative is 0 or UNDEFINED. They won't necessarily give us max or min values, but they are critical points, giving us some information :O
\[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1-x} }\cdot \frac{-1}{(1+x)^2}\]\[f'(x)=-\frac{1}{ \sqrt{1-x} \sqrt{1+x}}\]

Hmmm so we can't divide by 0, so -1 is out.
Anddd we can't take the square root of a negative number, meaning,
Anything smaller than -1 is out (it will make the bottom negative).
Anything larger than 1 is out (it will make the top negative).
I think our domain looks like this.
\[-1 < x \le 1\]

Automatically? :o hmm i dunno.
I just know that they're critical points because the first derivative told us so.
It turns out that this is probably one of those weird functions where we'll end up having a MINIMUM at x=1 even though it doesn't have a slope of 0 there :D

yah, f'(x)=0 will give us critical points (which there were none).
and also a and b are considered critical points.
But b wasn't in our domain, so we don't even consider that point.
Yah sounds good :D
So just the lonely little x=1.